The Differential of a Multivariable Function

Consider a differentiable function $f$ of two variables $z = f (x, y)$ . If $x$ changes to $x + Δ x$ and $y$ changes to $y + Δ y$ , the increment of $f$ , $Δ f$ , from the definition of differentiability , can be written as:
$\begin{aligned} Δ f & = f (x + Δ x, y + Δ y) - f (x, y) \\ (i) & = f_{x} (x, y) Δ x + f_{y} (x, y) Δ y + \sqrt{(Δ x)^{2} + (Δ y)^{2}} ε (Δ x, Δ y) . \end{aligned}$

We take the linear part of $Δ f$ and call them the differential of $f$ . The differential part of $f$ is denoted by $d f$ or $d z$ :
$\begin{aligned} d z = d f = & \frac{\partial f}{\partial x} Δ x + \frac{\partial f}{\partial y} Δ y \\ = & f_{x} (x, y) Δ x + f_{y} (x, y) Δ y . \end{aligned}$

Note that $d f$ is a function of four variables: $x, y, Δ x$ , and $Δ y$ . To emphasize on that, we may write it as $d f (x, y, Δ x, Δ y)$ .

If $x$ and $y$ are independent variables, from Equation (i) we have:
$d x = \underset{= 1}{\underset{⏟}{\frac{\partial x}{\partial x}}} Δ x + \underset{= 0}{\underset{⏟}{\frac{\partial x}{\partial y}}} Δ y = Δ x,$ and
$d y = \underset{= 0}{\underset{⏟}{\frac{\partial y}{\partial x}}} Δ x + \underset{= 1}{\underset{⏟}{\frac{\partial y}{\partial y}}} Δ y = Δ y .$

Then Equation (i) takes the form:
$d f = \frac{\partial f}{\partial x} d x + \frac{\partial f}{\partial y} d y = f_{x} (x, y) d x + f_{y} (x, y) d y .$
The above expression is sometimes called the total differential of $f (x, y)$ .

Definition 1. If $z = f (x, y)$ is a differentiable function at $(x, y)$ , the total differential of $f$ is the function $d f$ defined by:
$d f (x, y, d x, d y) = \frac{\partial f}{\partial x} d x + \frac{\partial f}{\partial y} d y = f_{x} (x, y) d x + f_{y} (x, y) d y$

When u = f(x, y, z):

Obviously, we can extend these methods and results to functions of any number of variables. For example, if $u = f (x, y, z)$ , then
$d u = d f = \frac{\partial f}{\partial x} d x + \frac{\partial f}{\partial y} d y + \frac{\partial f}{\partial z} d z .$

In general, when y = f(x₁, x₂, …, x_n):

In this case,

$d y = d f = \frac{\partial f}{\partial x_{1}} d x_{1} + \dots + \frac{\partial f}{\partial x_{n}} d x_{n} = \sum_{i = 1}^{n} \frac{\partial f}{\partial x_{i}} d x_{i} .$

Example 1

The period of vibration, $T$ of a mass on a spring is determined by the mass $m$ and the stiffness of the spring $k$ as: $T = 2 π \sqrt{\frac{m}{k}} .$ Estimate the percentage change in the period of this system if the mass increases by $5 %$ and the stiffness by $3 %$ .

Solution

We can use the total differential of $T$ and say $\frac{Δ T}{T} \approx \frac{d T}{T}$ :
$\begin{aligned} d T & = \frac{\partial T}{\partial m} d m + \frac{\partial T}{\partial k} d k \\ = 2 π \frac{1}{2 \sqrt{m k}} d m - \frac{2 π}{2} \sqrt{m} k^{- 3 / 2} d k, (because T = 2 π m^{1 / 2} k^{- 1 / 2}) \end{aligned}$
Here $m$ and $k$ are independent variables, so:
$d m = Δ m = \frac{5}{100} m, and d k = Δ k = \frac{3}{100} k$
If we plug these expressions for $d m$ and $d k$ in $d T$ , we obtain:
$\begin{aligned} d T = & \underset{\frac{5 π}{100} \sqrt{\frac{m}{k}}}{\underset{⏟}{π \frac{1}{\sqrt{m k}} \times \frac{5}{100} m}} - \underset{\frac{3 π}{100} \sqrt{\frac{m}{k}}}{\underset{⏟}{π \sqrt{m} k^{- 3 / 2} \times \frac{3}{100} k}} \\ = & - \frac{2 π}{100} \sqrt{\frac{m}{k}} \\ = & - \frac{1}{100} \times \underset{T}{\underset{⏟}{2 π \sqrt{\frac{m}{k}}}} \end{aligned}$
Therefore
$Δ T / T \approx d T / T = - 1 %$ , that is the period decreases by approximately 1%.

The exact change in the period is:
$\begin{aligned} Δ T & = 2 π \sqrt{\frac{m + 0.05 m}{k + 0.03 k}} - 2 π \sqrt{\frac{m}{k}} \\ = (\sqrt{\frac{1.05}{1.03}} - 1) \times 2 π \sqrt{\frac{m}{k}} \\ = (\sqrt{\frac{1.05}{1.03}} - 1) T \approx 0.009662 T . \end{aligned}$
That is, the exact change in the period is 0.9662%.